How much heat is
needed to heat 45 kg
of water from 18°C
to 83°C?

The Old Thermodynamist says:

Recall ΔH = m•Cp•ΔT where H is enthalpy (heat), m is mass Cp is heat capacity and ΔT is temperature change.

The heat capacity of water is dependent on temperature, but is commonly taken as 4.186kJ/kg-K

In this case, we have 8kg of water and a temperature change ΔT = 80K

So ΔH = 8(4.186)(80) = 2679.0kJ

The heat required is 2679.0kJ (which is 0.744kWh

Explanation:

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in this image, you can see solar radiation going through ocean water.

no image, but probably absorption